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Physics 322A, Winter 2018 Exam 1, page 5 ED-UWA322A181T-E1(AML)Sol.docx

IV. [25 points total] Tutorial question.

A non-magnetic slab with uniform current density Jo directed out of the page (−��) fills the space from z = −d to z = d. The slab is infinite in both x and y. The magnetic field at the origin (Point O) is zero. There are no other sources in this system.

A. [7 pts.] What component(s) of the magnetic field due to this slab at point P must be zero? For each component you identified, explain your reasoning.

y-component: Biot-Savart Law shows that every dB from every small current element must be perpendicular to that current element. Since all of the current elements only point in y, the magnetic field cannot have a component in that direction. z-component: A 180° rotational symmetry about any y-axis along z = 0 shows that a magnetic field with a z-component above the slab must have the opposite z-component below the slab. This is inconsistent with Ñ×B = 0, so the magnetic field cannot have a z-component. One could also argue that the superposition of the z-component, as wires from the right side of the slab will have a downward component while wires from the left side will have an equally strong upward component.

B. [10 pts.] Consider an open, semi-circular arc of radius 𝑟𝑜 < 𝑑 centered around the origin, as shown at right. Determine the line integral ∫ �� ∙ 𝑑𝑙 𝑅𝐿 . Explain your reasoning and/or show your work.

First of all, you don’t need to evaluate �� , but the integral ∫ �� ∙ 𝑑𝑙 𝑅𝐿 . So

although the path is poorly constructed in that the direction is not with �� nor is �� constant along the path, that’s not important as you do not need to simplify the integral to pull B out. That being said, the main problem is that Ampere’s Law as ∮ �� ∙ 𝑑𝑙 = 𝜇𝑜 ∬ 𝐽 ∙ 𝑑𝑎 = 𝜇𝑜𝐼𝑒𝑛𝑐 can only be used for a loop that is closed. So to fix this issue, one can close the loop! The two simplest closed loops involve adding a straight path from R to the origin to L, or completing the circle clockwise. In the first case, the added path is in a region where B = 0, with the given that B is zero at the origin with x-translational symmetry. In the second case the bottom part’s line integral will be equivalent to the top part’s line integral because of the discrete rotational symmetry about the y-axis through the origin.

So after closing the loop, one can find the current encircled (and divide by 2 if using the full circle). Both methods will lead to a line integral of − 𝜇𝑜Jo𝜋𝑟𝑜2

2. Negative sign because the path is going in the

“wrong” direction.

C. [8 pts] Consider the magnitude of the magnetic field for points inside the slab.

How does the magnitude of the magnetic field depend on y? Explain.

Constant in y: Continuous translational symmetry along the y-axis. How does the magnitude of the magnetic field depend on z? Explain.

Linear in z: Looking at ∮ �� ∙ 𝑑𝑙 = 𝜇𝑜 ∬ 𝐽 ∙ 𝑑𝑎 = 𝜇𝑜𝐼𝑒𝑛𝑐, moving the top part of a rectangular loop up by Δz increases the current enclosed by a factor of Δz. dl is horizontal so it remains constant, which means the factor of Δz must come from the magnitude of the magnetic field. You can also look at the Cartesian form of ∇ × �� = 𝜇𝑜𝐽 and argue that with symmetry arguments from part A and above (with the x-dependence constant as well), the only term is the 𝑑𝐵𝑥

𝑑𝑧 term. Since this derivative is

constant, the field must be linear in z.

Jo

Infinite slab of current

O

P

x

z

y

z = d

z = −d

O

z = d

z = −d

RLro